```html
<!DOCTYPE html>
<html lang="zh-CN">
<head>
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    <title>算法可视化：两数相加</title>
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</head>
<body>
    <!-- Hero Section -->
    <div class="hero-gradient text-white py-20 px-4 sm:px-6 lg:px-8">
        <div class="max-w-4xl mx-auto text-center">
            <h1 class="text-4xl md:text-5xl font-bold mb-6 font-serif">两数相加算法解析</h1>
            <p class="text-xl md:text-2xl opacity-90 mb-8 max-w-3xl mx-auto">
                深入理解链表操作与加法模拟的高效实现
            </p>
            <div class="flex justify-center space-x-4">
                <div class="flex items-center">
                    <i class="fas fa-link text-xl mr-2"></i>
                    <span>链表操作</span>
                </div>
                <div class="flex items-center">
                    <i class="fas fa-calculator text-xl mr-2"></i>
                    <span>数学运算</span>
                </div>
                <div class="flex items-center">
                    <i class="fas fa-project-diagram text-xl mr-2"></i>
                    <span>算法模拟</span>
                </div>
            </div>
        </div>
    </div>

    <!-- Main Content -->
    <div class="max-w-6xl mx-auto px-4 sm:px-6 lg:px-8 py-12">
        <!-- Problem Description -->
        <section class="mb-16">
            <div class="flex items-center mb-6">
                <div class="w-12 h-12 rounded-full bg-indigo-100 flex items-center justify-center mr-4">
                    <i class="fas fa-question text-indigo-600 text-xl"></i>
                </div>
                <h2 class="text-2xl md:text-3xl font-bold text-gray-800">题目描述</h2>
            </div>
            <div class="bg-white rounded-xl shadow-md p-6 md:p-8 transition duration-300 card-hover">
                <p class="text-gray-700 mb-4 text-lg leading-relaxed">
                    给定两个非空链表，表示两个非负整数。它们的每个节点存储一位数字，且是<span class="font-bold text-indigo-600">逆序存储</span>（个位在链表头）。将这两个数相加，并以链表形式返回。
                </p>
                <p class="text-gray-700 text-lg leading-relaxed">
                    例如，输入 <span class="font-mono bg-gray-100 px-2 py-1 rounded">(2 → 4 → 3) + (5 → 6 → 4)</span>，表示 <span class="font-bold">342 + 465 = 807</span>，返回 <span class="font-mono bg-gray-100 px-2 py-1 rounded">7 → 0 → 8</span>。
                </p>
            </div>
        </section>

        <!-- Visualization Section -->
        <section class="mb-16">
            <div class="flex items-center mb-6">
                <div class="w-12 h-12 rounded-full bg-purple-100 flex items-center justify-center mr-4">
                    <i class="fas fa-project-diagram text-purple-600 text-xl"></i>
                </div>
                <h2 class="text-2xl md:text-3xl font-bold text-gray-800">算法可视化</h2>
            </div>
            <div class="bg-white rounded-xl shadow-md p-6 md:p-8 mb-8 transition duration-300 card-hover">
                <div class="flex flex-col items-center">
                    <h3 class="text-xl font-semibold mb-6 text-gray-800">链表相加过程演示</h3>
                    
                    <div class="flex justify-center items-center mb-10">
                        <!-- List 1 -->
                        <div class="flex items-center mr-8">
                            <div class="node bg-indigo-100 text-indigo-700 border-2 border-indigo-300">2</div>
                            <div class="arrow"></div>
                            <div class="node bg-indigo-100 text-indigo-700 border-2 border-indigo-300">4</div>
                            <div class="arrow"></div>
                            <div class="node bg-indigo-100 text-indigo-700 border-2 border-indigo-300">3</div>
                        </div>
                        
                        <div class="text-3xl font-bold text-gray-500 mr-8">+</div>
                        
                        <!-- List 2 -->
                        <div class="flex items-center mr-8">
                            <div class="node bg-purple-100 text-purple-700 border-2 border-purple-300">5</div>
                            <div class="arrow"></div>
                            <div class="node bg-purple-100 text-purple-700 border-2 border-purple-300">6</div>
                            <div class="arrow"></div>
                            <div class="node bg-purple-100 text-purple-700 border-2 border-purple-300">4</div>
                        </div>
                        
                        <div class="text-3xl font-bold text-gray-500 mr-8">=</div>
                        
                        <!-- Result -->
                        <div class="flex items-center">
                            <div class="node bg-green-100 text-green-700 border-2 border-green-300">7</div>
                            <div class="arrow"></div>
                            <div class="node bg-green-100 text-green-700 border-2 border-green-300">0</div>
                            <div class="arrow"></div>
                            <div class="node bg-green-100 text-green-700 border-2 border-green-300">8</div>
                        </div>
                    </div>
                    
                    <div class="text-center max-w-2xl">
                        <div class="bg-blue-50 border-l-4 border-blue-500 p-4 mb-6">
                            <p class="text-blue-800"><i class="fas fa-info-circle mr-2"></i>算法从链表头部开始逐位相加，处理进位，最后返回新链表</p>
                        </div>
                    </div>
                </div>
            </div>
        </section>

        <!-- Algorithm Explanation -->
        <section class="mb-16">
            <div class="flex items-center mb-6">
                <div class="w-12 h-12 rounded-full bg-blue-100 flex items-center justify-center mr-4">
                    <i class="fas fa-brain text-blue-600 text-xl"></i>
                </div>
                <h2 class="text-2xl md:text-3xl font-bold text-gray-800">核心算法解析</h2>
            </div>
            <div class="grid md:grid-cols-2 gap-8">
                <div class="bg-white rounded-xl shadow-md p-6 md:p-8 transition duration-300 card-hover">
                    <h3 class="text-xl font-semibold mb-4 text-gray-800 flex items-center">
                        <i class="fas fa-bullseye mr-2 text-indigo-500"></i>核心考点
                    </h3>
                    <ul class="space-y-3 text-gray-700">
                        <li class="flex items-start">
                            <i class="fas fa-check-circle text-green-500 mt-1 mr-2"></i>
                            <span>链表操作：理解链表节点的遍历和连接</span>
                        </li>
                        <li class="flex items-start">
                            <i class="fas fa-check-circle text-green-500 mt-1 mr-2"></i>
                            <span>数学运算：处理加法进位和数字位运算</span>
                        </li>
                        <li class="flex items-start">
                            <i class="fas fa-check-circle text-green-500 mt-1 mr-2"></i>
                            <span>边界条件：处理不同长度链表和最后的进位</span>
                        </li>
                    </ul>
                </div>
                
                <div class="bg-white rounded-xl shadow-md p-6 md:p-8 transition duration-300 card-hover">
                    <h3 class="text-xl font-semibold mb-4 text-gray-800 flex items-center">
                        <i class="fas fa-lightbulb mr-2 text-yellow-500"></i>解题思路
                    </h3>
                    <ol class="space-y-3 text-gray-700 list-decimal list-inside">
                        <li>使用哑节点(dummy node)简化链表操作</li>
                        <li>同时遍历两个链表，逐位相加，处理进位</li>
                        <li>考虑链表长度不一致的情况</li>
                        <li>最后检查是否有剩余的进位需要处理</li>
                    </ol>
                </div>
            </div>
        </section>

        <!-- Complexity Analysis -->
        <section class="mb-16">
            <div class="flex items-center mb-6">
                <div class="w-12 h-12 rounded-full bg-red-100 flex items-center justify-center mr-4">
                    <i class="fas fa-chart-line text-red-600 text-xl"></i>
                </div>
                <h2 class="text-2xl md:text-3xl font-bold text-gray-800">复杂度分析</h2>
            </div>
            <div class="grid md:grid-cols-2 gap-8">
                <div class="bg-white rounded-xl shadow-md p-6 md:p-8 transition duration-300 card-hover">
                    <h3 class="text-xl font-semibold mb-4 text-gray-800 flex items-center">
                        <i class="fas fa-clock mr-2 text-purple-500"></i>时间复杂度
                    </h3>
                    <p class="text-gray-700">
                        O(max(n, m))，其中n和m分别是两个链表的长度。算法需要遍历两个链表的全部节点。
                    </p>
                    <div class="mt-4">
                        <div class="bg-purple-50 p-4 rounded-lg">
                            <p class="text-purple-800 font-medium"><i class="fas fa-star mr-2"></i>最优情况：已经是最优解法</p>
                        </div>
                    </div>
                </div>
                
                <div class="bg-white rounded-xl shadow-md p-6 md:p-8 transition duration-300 card-hover">
                    <h3 class="text-xl font-semibold mb-4 text-gray-800 flex items-center">
                        <i class="fas fa-memory mr-2 text-blue-500"></i>空间复杂度
                    </h3>
                    <p class="text-gray-700">
                        O(max(n, m))，新链表的长度最多为max(n,m)+1（因为有进位）。
                    </p>
                    <div class="mt-4">
                        <div class="bg-blue-50 p-4 rounded-lg">
                            <p class="text-blue-800 font-medium"><i class="fas fa-star mr-2"></i>节省空间：只创建必要的节点</p>
                        </div>
                    </div>
                </div>
            </div>
        </section>

        <!-- Code Implementation -->
        <section class="mb-16">
            <div class="flex items-center mb-6">
                <div class="w-12 h-12 rounded-full bg-green-100 flex items-center justify-center mr-4">
                    <i class="fas fa-code text-green-600 text-xl"></i>
                </div>
                <h2 class="text-2xl md:text-3xl font-bold text-gray-800">代码实现</h2>
            </div>
            <div class="bg-white rounded-xl shadow-md overflow-hidden transition duration-300 card-hover">
                <div class="flex items-center justify-between bg-gray-800 px-4 py-3">
                    <div class="flex items-center">
                        <div class="w-3 h-3 rounded-full bg-red-500 mr-2"></div>
                        <div class="w-3 h-3 rounded-full bg-yellow-500 mr-2"></div>
                        <div class="w-3 h-3 rounded-full bg-green-500"></div>
                    </div>
                    <span class="text-gray-300 text-sm">Python 实现</span>
                </div>
                <div class="code-block p-4 overflow-x-auto">
                    <pre class="text-gray-200 font-mono text-sm md:text-base">
<span class="text-pink-400">def</span> <span class="text-yellow-300">addTwoNumbers</span>(l1, l2):
    <span class="text-gray-400"># 创建哑节点简化操作</span>
    dummy = <span class="text-blue-400">ListNode</span>(<span class="text-orange-400">0</span>)
    curr = dummy
    carry = <span class="text-orange-400">0</span>
    
    <span class="text-pink-400">while</span> l1 <span class="text-pink-400">or</span> l2 <span class="text-pink-400">or</span> carry:
        <span class="text-gray-400"># 获取当前位的值，如果链表已经结束则为0</span>
        x = l1.val <span class="text-pink-400">if</span> l1 <span class="text-pink-400">else</span> <span class="text-orange-400">0</span>
        y = l2.val <span class="text-pink-400">if</span> l2 <span class="text-pink-400">else</span> <span class="text-orange-400">0</span>
        
        <span class="text-gray-400"># 计算和与进位</span>
        total = x + y + carry
        carry = total // <span class="text-orange-400">10</span>
        
        <span class="text-gray-400"># 创建新节点并移动指针</span>
        curr.next = <span class="text-blue-400">ListNode</span>(total % <span class="text-orange-400">10</span>)
        curr = curr.next
        
        <span class="text-gray-400"># 移动链表指针</span>
        l1 = l1.next <span class="text-pink-400">if</span> l1 <span class="text-pink-400">else</span> <span class="text-blue-400">None</span>
        l2 = l2.next <span class="text-pink-400">if</span> l2 <span class="text-pink-400">else</span> <span class="text-blue-400">None</span>
    
    <span class="text-gray-400"># 返回结果链表的头节点</span>
    <span class="text-pink-400">return</span> dummy.next</pre>
                </div>
            </div>
        </section>

        <!-- Key Points -->
        <section>
            <div class="flex items-center mb-6">
                <div class="w-12 h-12 rounded-full bg-yellow-100 flex items-center justify-center mr-4">
                    <i class="fas fa-key text-yellow-600 text-xl"></i>
                </div>
                <h2 class="text-2xl md:text-3xl font-bold text-gray-800">关键点解析</h2>
            </div>
            <div class="grid md:grid-cols-2 gap-8">
                <div class="bg-white rounded-xl shadow-md p-6 md:p-8 transition duration-300 card-hover">
                    <h3 class="text-xl font-semibold mb-4 text-gray-800 flex items-center">
                        <i class="fas fa-check-circle mr-2 text-green-500"></i>哑节点的使用
                    </h3>
                    <p class="text-gray-700 mb-4">
                        哑节点(dummy node)简化了链表操作，避免了对头节点的特殊处理。创建一个临时节点作为结果链表的起始点，最后返回它的下一个节点。
                    </p>
                    <div class="flex items-center text-sm text-gray-500">
                        <i class="fas fa-code mr-2"></i>
                        <code class="bg-gray-100 px-2 py-1 rounded">dummy = ListNode(0)</code>
                    </div>
                </div>
                
                <div class="bg-white rounded-xl shadow-md p-6 md:p-8 transition duration-300 card-hover">
                    <h3 class="text-xl font-semibold mb-4 text-gray-800 flex items-center">
                        <i class="fas fa-check-circle mr-2 text-green-500"></i>进位处理
                    </h3>
                    <p class="text-gray-700 mb-4">
                        进位(carry)是关键，每次相加都要计算当前位的值和新的进位。循环条件包含carry确保最后的进位不被遗漏。
                    </p>
                    <div class="flex items-center text-sm text-gray-500">
                        <i class="fas fa-code mr-2"></i>
                        <code class="bg-gray-100 px-2 py-1 rounded">carry = total // 10</code>
                    </div>
                </div>
                
                <div class="bg-white rounded-xl shadow-md p-6 md:p-8 transition duration-300 card-hover">
                    <h3 class="text-xl font-semibold mb-4 text-gray-800 flex items-center">
                        <i class="fas fa-check-circle mr-2 text-green-500"></i>链表遍历
                    </h3>
                    <p class="text-gray-700 mb-4">
                        使用while循环同时遍历两个链表，当链表长度不一致时，较短链表的空位视为0。循环条件确保所有位都被处理。
                    </p>
                    <div class="flex items-center text-sm text-gray-500">
                        <i class="fas fa-code mr-2"></i>
                        <code class="bg-gray-100 px-2 py-1 rounded">while l1 or l2 or carry:</code>
                    </div>
                </div>
                
                <div class="bg-white rounded-xl shadow-md p-6 md:p-8 transition duration-300 card-hover">
                    <h3 class="text-xl font-semibold mb-4 text-gray-800 flex items-center">
                        <i class="fas fa-check-circle mr-2 text-green-500"></i>边界条件
                    </h3>
                    <p class="text-gray-700 mb-4">
                        需要处理链表长度不同和最后有进位的情况。例如：5+5=10，需要创建两个节点(0→1)而不是一个节点(0)。
                    </p>
                    <div class="flex items-center text-sm text-gray-500">
                        <i class="fas fa-code mr-2"></i>
                        <code class="bg-gray-100 px-2 py-1 rounded">curr.next = ListNode(total % 10)</code>
                    </div>
                </div>
            </div>
        </section>
    </div>

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```